Superposition and Equilibrium

Many charges at once

Coulomb's law is for two charges. Real problems have three, four or many. Experiment gives a simple rule. The force that one charge exerts on another does not change when more charges are brought nearby. So each pair can be treated alone, and the forces on any one charge then add as vectors. This is the principle of superposition, the same rule you used for gravitational forces.

The work comes in three steps. Find the size of each force from Coulomb's law. Draw its direction from the signs: away from a like charge, towards an unlike charge. Then add the arrows, along one line by signed numbers, in a plane by components.

In a plane the arrows point in different directions, so add them by components. Put \(q_0 = +1\,\mu\text{C}\) at the origin, \(q_1 = +1\,\mu\text{C}\) on the \(x\) axis at \(15\,\text{cm}\), and \(q_2 = -3\,\mu\text{C}\) on the \(y\) axis at \(30\,\text{cm}\). Then \(q_1\) pushes \(q_0\) along \(-x\) with \(0.4\,\text{N}\). And \(q_2\) pulls it along \(+y\) with \(0.3\,\text{N}\). The two forces are at right angles, so the net force is \(\sqrt{0.4^2 + 0.3^2}\) \(= 0.5\,\text{N}\), at \(37°\) from the \(-x\) direction towards \(+y\).

Charges at the corners of a triangle and a square

Place \(2\,\mu\text{C}\) at each corner of an equilateral triangle of side \(20\,\text{cm}\). Take the charge at one corner. The other two push it with \(9 \times 10^9 \times (2 \times 10^{-6})^2/0.04\) \(= 0.9\,\text{N}\) each. The two pushes are \(60°\) apart, so their sum is \(\sqrt{F^2 + F^2 + 2F^2\cos 60°}\) \(= \sqrt3\,F\) \(= 1.56\,\text{N}\). By symmetry it points along the bisector of that corner, away from the triangle.

A square works the same way, with one more step. Put \(1\,\mu\text{C}\) at each corner of a square of side \(10\,\text{cm}\), and take the charge at corner A. The two neighbours push with \(0.9\,\text{N}\) each, at right angles. Their sum is \(0.9\sqrt2 = 1.27\,\text{N}\), along the diagonal. The far corner is \(10\sqrt2\,\text{cm}\) away. It pushes along the same diagonal with half the force, \(0.45\,\text{N}\). So the net force is \(1.27 + 0.45 = 1.72\,\text{N}\), along the diagonal away from the square.

Symmetry at the centre of a polygon

Six equal charges \(q\) sit at the corners of a regular hexagon, each a distance \(a\) from a charge \(q_0\) at the centre. Each corner pushes \(q_0\) with the same force \(kqq_0/a^2\), straight away from itself. Opposite corners push in opposite directions. So the six forces cancel in pairs, and the net force is zero. For a regular pentagon there are no opposite pairs, yet the five forces still cancel. Turn the pentagon by one corner and it looks exactly the same, so the net force cannot point any particular way; it must be zero.

The removed-corner trick is worth remembering. The full polygon gives zero, so the charges that remain give exactly the negative of the missing one. The same trick works when one corner is reversed. Turning \(+q\) into \(-q\) is the same as adding \(-2q\) at that corner. So the net force becomes \(2kqq_0/a^2\), directed towards the reversed charge, as the third chip shows.

The null point

Put \(+4\,\mu\text{C}\) and \(+1\,\mu\text{C}\) \(30\,\text{cm}\) apart. Somewhere on the line a third charge must feel equal and opposite pushes from the two. Such a place is a null point. Let it be \(x\) from the \(4\,\mu\text{C}\) charge. Equal forces need \(4/x^2 = 1/(0.3 - x)^2\). Taking square roots, \(2(0.3 - x) = x\), so \(x = 0.2\,\text{m}\). The null point is \(20\,\text{cm}\) from the larger charge, nearer the smaller one.

The third charge \(q\) appears on both sides of the balance and cancels. So the null point is the same for a positive or a negative third charge, large or small. It is a property of the two fixed charges alone.

Now make the second charge unlike, \(-1\,\mu\text{C}\). Between the two charges both forces on a third charge point the same way, so they cannot cancel there. Beyond the larger charge the larger charge is also nearer, so it always wins. The null point must lie outside, beyond the smaller charge. With \(x\) from the \(+4\,\mu\text{C}\), \(4/x^2 = 1/(x - 0.3)^2\) gives \(x = 0.6\,\text{m}\). That is \(30\,\text{cm}\) beyond the \(-1\,\mu\text{C}\) charge, as the figure above shows.

Three charges all at rest

Two charges of \(+4\,\mu\text{C}\) sit at \(x = 0\) and \(x = 40\,\text{cm}\). Where can a third charge rest, and which third charge would keep all three at rest? The third charge feels no force only at the midpoint, \(20\,\text{cm}\) from each. Now look at the charge at \(x = 0\). The other \(+4\,\mu\text{C}\) pushes it away with \(k(4)(4) \times 10^{-12}/0.16\). The middle charge must pull it back just as hard. So it is negative, and \(k(4)|q| \times 10^{-12}/0.04\) must equal that push. This gives \(|q| = 1\,\mu\text{C}\): a charge of \(-1\,\mu\text{C}\) at the midpoint.

This is the method of the chapter Objects in Equilibrium, applied to electric forces: the net force on every body is zero. Find where the middle charge must sit, from its own balance. Then find how large it must be, from the balance of one end charge.

Charged balls on threads

Two identical balls hang from one point by threads and carry equal charges. They push each other apart until each thread leans at an angle \(\theta\) to the vertical. Three forces act on each ball: its weight \(mg\) down, the tension \(T\) along the thread, and the electric push \(F\) horizontally. Draw the free-body diagram exactly as in the chapter Objects in Equilibrium.

Resolve along and across the vertical. Vertically, \(T\cos\theta = mg\). Horizontally, \(T\sin\theta = F\). Dividing the second by the first gives \(F = mg\tan\theta\). So the angle alone tells you the electric force, and Coulomb's law then gives the charge.